Shuangshuang · 2020年09月25日
问题如下：
When modeling using a time trend model, what is the relationship between and for any forecasting period h? Are these ever the same? Assume the error terms is normally distributed around a mean of zero.
解释：
A time trend model for can be stated as:
,
where g(t) is a function of t.
So,
,
which gives
,
On the other hand:
,
which equals
And so:
These will be equal if the variance is zero (in other words, if the process is completely deterministic.
误差服从idd，那误差的指数怎么求方差啊？能否讲一下sigma平方/2怎么来的呀？
NO.PZ2020011101000019问题如下When moling lnYtln Y_tlnYt using a time trenmol, whis the relationship between expET[lnYT+h]exp E_T[ln Y_{T+ h}]expET[lnYT+h] anET[YT+h]E_T[Y_{T+ h}]ET[YT+h] for any forecasting perioh? Are these ever the same? Assume the error terms is normally stributearouna meof zero. A time trenmol for lnYtln Y_tlnYt cstateas: lnYt=g(t)+ϵt,ϵ∼N(0,σ2)ln Y_t = g(t) + \epsilon_t, \epsilon ∼ N(0, \sigma^2)lnYt=g(t)+ϵt,ϵ∼N(0,σ2),where g(t) is a function of t. So,ET[lnYT+h]=g(T+h)E_T[ln Y_{T+ h}] = g(T + h)ET[lnYT+h]=g(T+h), whigivesexpET[lnYT+h]=exp[g(T+h)]exp E_T[ln Y_{T+ h}] = exp [g(T + h)]expET[lnYT+h]=exp[g(T+h)],On the other hanET[YT+h]=ET[exp(g(T+h)+ϵT+h)]=exp(g(T+h)+ET[exp epsilonT+h)]E_T[Y_{T+ h}] = E_T[exp(g(T + h) + \epsilon_{T+ h})] = exp(g(T + h) + E_T[exp \ epsilon_{T+ h})]ET[YT+h]=ET[exp(g(T+h)+ϵT+h)]=exp(g(T+h)+ET[exp epsilonT+h)],whiequals ET[YT+h]=exp[[g(T+h)+σ2/2]E_T\lbraY_{T+h}\rbrack=exp\left[\lbrag(T+h)+\sigma^2/2\right]ET[YT+h]=exp[[g(T+h)+σ2/2]Anso: ET[YT+h]=expET[[lnYT+h+σ2/2]E_T\lbraY_{T+h}\rbrack=expE_T\left[\lbralnY_{T+h}+\sigma^2/2\right]ET[YT+h]=expET[[lnYT+h+σ2/2]These will equif the varianis zero (in other wor, if the process is completely terministi老师好，1、红线等式看不明白，是怎么等价过来的呢？2、蓝色线是为什么相等呢？能给个推导过程吗？
NO.PZ2020011101000019 老师，这道题的考点是啥？需要掌握哪个知识点呢？
NO.PZ2020011101000019 E(exp(εt+h)))为什么不等于1而是σ^2呀？E（εt+h)不是等于0吗？谢谢！
NO.PZ2020011101000019 请问这步是怎么得到的？ET[exp(g(T+h)+ϵT+h)]=exp(g(T+h)+ET[exp epsilonT+h)] 以及后一项如何转化为下一步的sigma^2/2？ 谢谢！
